Competitive exams have at least one question where you have to find the square root of a number in seconds. With the help of this **shortcut method, **you can **find square root of a number** within seconds. This method is applicable only for perfect squares. The Square root questions asked in competitive exams/bank exams & entrance exams are usually perfect squares. Refer How to check if a number is a perfect square? to find out if a given number is a perfect square.

## Divisibility Rules – Number divisible by 2,3,4,5,6,7,8,9,10,11 or 13

Divisibility Rules are useful when you want to find out if the given big number is exactly divisible or not without actually dividing.

This saves a lot of time which is the need of the hour in any competitive exam. This shortcut is especially handy when you are solving Problems on Ages. We shall find out if a number is divisible by 2,3,4,5,6,7,8,9,10,11,13 using various divisibility rules.

## IIBF Courses, Certification, Exams 2022

IIBF offers different courses for Banking and Finance Professionals to excel in their field. The Indian Institute of Banking and Finance (IIBF) conducts exams twice a year in May/ June and November/ December.

## Sum of Consecutive Numbers Shortcut

Finding the sum of consecutive numbers is a common question asked in competitive exams. You can now find the sum of consecutive numbers starting with 1, the sum of consecutive odd numbers, and the sum of consecutive even numbers in seconds.

**What are consecutive numbers?**

Numbers that follow each other in order are Consecutive numbers. Here the difference between every two numbers is 1. The consecutive numbers can be represented as n,n+1,n+2.

Examples of consecutive numbers:

(1)1,2,3,4,5,6,7,8

(2)6,7,8,9,10,11

**Sum of consecutive numbers starting from 1 shortcut**

Now that we know what consecutive numbers are. We shall see a shortcut to find the sum of consecutive numbers starting from 1.

It can be represented as sum of n consecutive numbers = n+(n+1)+(n+2)

**Sum of consecutive numbers starting from 1 shortcut steps**

Step 1:Count the total number of integers from 1 to the last consecutive number given.

Step 2:Multiply the result in step 1 by 1 more than the number.

Step 3:Divide the result of step 2 by 2.

We shall see some examples to understand better

Example 1:Find the sum of all consecutive numbers from 1 to 9.

Step 1: Counting the number of integers from 1 to 9. We find that there are 9 integers.

Step2: Multiplying the number of integers(9) by 1 more than the number that is 10.

We get, 9 X 10=90

Step3: Dividing the result obtained in step2 by 2

we get, 90/2=45

Ans Hence, the sum of consecutive numbers from 1 to 9=45

Example 2:Find the sum of all consecutive numbers from 1 to 99.

Step 1: Counting the number of integers from 1 to 99. We find that there are 99 integers.

Step2: Multiplying the number of integers(99) by 1 more than the number that is 100.

We get, 99 X 100=9900

Step3: Dividing the result obtained in step2 by 2

we get, 9900/2=4950

Ans Hence, sum of consecutive numbers from 1 to 99=4,950

**Odd Consecutive numbers**

Odd numbers that follow each other are consecutive odd numbers. Here the difference between every two odd numbers is 2.

It can be represented as Sum of n numbers = n+(n+2)+(n+4)+(n+6) where n is an odd number..

We shall see how to add odd consecutive numbers.

### Sum of consecutive odd numbers starting from 1 steps

Step 1: Count the total number of odd integers from 1 to the last consecutive number given.

Step 2: Square the result obtained in step 1.

Example 1:Find the sum of all consecutive odd numbers from 1 to 10.

Step 1: Counting the number of integers from 1 to 9. We find that there are 5 integers.

Step 2: Squaring the result obtained in step 1.

We get, 5^{2}=25

Ans hence, Sum of odd numbers from 1 to 9 is 25

Example 2:Find the sum of all consecutive odd numbers from 1 to 100.

Step 1: Counting the number of integers from 1 to 100. We find that there are 50 integers.

Step 2: Squaring the result obtained in step 1.

We get, 50^{2}=2500

Ans 2500

**Even Consecutive numbers**

Even numbers that follow each other are consecutive even numbers. Here the difference between every two even numbers is 2.

It can be represented as Sum of n numbers = n+(n+2)+(n+4)+(n+6)where n is an even number.

We shall see how to add even consecutive numbers.

### Sum of consecutive even numbers starting from 2 steps

Step 1: Count the total number of odd integers from 1 to the last consecutive number given.

Step 2: Multiply the result in step 1 by 1 more than the number.

Example 1:Find the sum of all consecutive even numbers from 1 to 10.

Step 1: Counting the number of integers from 1 to 10. We find that there are 5 integers.

Step 2: Multiplying the number of integers(5) by 1 more than the number that is 6.

We get, 5 X 6=30

Ans 30

Example 2:Find the sum of all consecutive even numbers from 1 to 100.

Step 1: Counting the number of integers from 1 to 100. We find that there are 50 integers.

Step 2: Multiplying the number of integers(50) by 1 more than the number that is 51.

We get, 50 X 51=2550

Ans 2550.

## Upcoming Bank Exams 2021 [Updated Now]

Upcoming bank exams for 2021 are listed here. With more than 200+ banks operating in India, the banking sector is one of the largest providers of jobs in the country. Lakhs of people take up individual bank exams conducted by banks and bank exams conducted by IBPS. IBPS conducts bank exams throughout the year for various positions across the country. Institute of Banking Personnel Selection (IBPS) conducts online exams recruitment in clerical posts, probationary officers, technical officers, etc. Be the first to know the registration date for a bank. Here you get all the information related to eligibility criteria for a particular job vacancy. Check to see if you meet the basic eligibility criteria and apply before the last date. Now, you will not miss out on any job opportunity. Here we provide you last date to apply for a post for all the upcoming basis.